Trying to simplify the expression $A'B'C'D' + A'BC'D' + A'BC'D + AB'C'D$

Question

So far I've got:

$A'B'C'D' + A'BC'D' + A'BC'D + AB'C'D$

$= A'C'D'(B' + B) + C'D(A'B + AB')$

$= A'C'D'(1) + C'D(A \;\text{ XOR }\; B)$

$= C'[A'D' + D(A \;\text{ XOR }\; B)]$

Did I do this correctly? Is there a simpler solution?

Thanks

K

Answer 1

Yes, your work is correct.

Simplification of a Boolean expression depends on context, and what form you are seeking in your "simplification": for example, conjunctive normal form (product of sums) or disjunctive normal form (sum of products), etc. Typically, one does not introduce "xor" $\oplus$ unless expressing the entire function in terms of $\oplus$, $\land$, $'$.

See, for example, the following, without the use of $\oplus$ (xor):

$$\begin{align} A'C'D'(B' + B) + A'BC'D + AB'C'D &= A'C'D' + A'BC'D + AB'C'D \\ \\ &= C'[A'D' + D(A'B + AB')]\end{align}$$

But again, all your manipulations are indeed correct.